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Previous year question hub

Hyperbola - Coordinate Geometry - Mathematics Previous Year Questions

Practice Hyperbola - Coordinate Geometry - Mathematics previous year questions organised from real papers, with year-wise coverage and clear topic navigation.

2Papers
2Years
2Questions
1Topics

Hyperbola question pattern

Every graph below is calculated only from this selection.

Questions by year

Year-wise coverage for Hyperbola. Each bar uses a separate theme-derived color.

Difficulty distribution

How the classified questions are distributed by difficulty.

Not classified 2 100%

Question type distribution

MCQ, numerical, multiple-select and other formats found in these papers.

Multiple Choices 2 100%

Subject weightage

Top subjects by unique question coverage.

Mathematics
2 Qs

Most asked topics

Top topics across the included previous year papers.

Coordinate Geometry
2 Qs

Subtopic coverage

Top subtopics inside this exact selection.

Hyperbola
2 Qs

Paper coverage

Question coverage for the most populated papers. Every active PYP paper remains listed below.

VITEEE 2024
1 Qs
VITEEE 2022
1 Qs

Included previous year papers

Newest papers appear first. Sort by year, question coverage or name.

PaperYear / sessionQuestions in this viewOpen
VITEEE 202420241View paper
VITEEE 202220221View paper

All Hyperbola previous year questions

Practice every matching question in batches of 20, with every available option.

1
2022 · Mathematics · Coordinate Geometry · Hyperbola
VITEEE 2022

Through a fixed point \(P(\alpha, \beta), a\), variable line is drawn to cut the coordinate axes at \(A\) and \(B\). The locus of the mid-point of \(A B\) is

A
a hyperbola with eccentricity 2
B
a hyperbola with centre \(\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)\)
C
a hyperbola with asymptotes along axes
D
not a hyperbola
Open complete paper
2
2024 · Mathematics · Coordinate Geometry · Hyperbola
VITEEE 2024

If $e$ is the eccentricity of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\theta$ is the angle between the asymptotes, then $\cos \frac{\theta}{2}$ is

A
$\frac{1}{e}$
B
$\frac{1}{\sqrt{e}}$
C
$-\frac{1}{e}$
D
$\frac{1}{\sqrt{e^2+2}}$
Open complete paper